Question

(1 II. Find eigen values and eigen vectors of A=0 LO 6. 0 2 -3 01 3 2)

Answer 1

I find eigen values to sepaonuofla ? O O AE 0 2 3 6 - 3 solm? then charecterstic et eigenvalue of a equ | A-NII 0 -0 => 2-d 0 3 २-d 0 - 3 2- 3 (P-1) = -O - 3 O >> 11-1) 112-12 + 9) => (4-1) lla d-)? +9] = (N-1) =0 (0-2)*+9 = +9 = 0 لان (1-2) = -2 d = 1 + 31 da! 'T-P PE+ 31.

ill eigen vector corresponding to d = 1 (A-I) = 0 0 x, 0 -> 3 x2 X3 -3 So we have a homogeneous it ly system Gaussian of linear Elimination equi SO we solve 3 R, R2 O 0 - 3 0 -3 -3 - R3+3R, 0 3 3 - 0 وه می REP 10 0 O 0 lo 0 12-31 R2/10 3 O R,R, 3R 0 0 O

So 2 ER 23 11 2, x, => 22 o 11 x, => X = x2 co 23 if 13 x - 1 then eigenvector v = iii eigznector corresponding to dz = 2-3i (A-(2-30) 1) -O 0 -1 + 30 X 3 xa Ooo O 30 X3 31 - 3

By Gaussian Elimination - + 0 0 o R, → R,/[-+31) / 3i 3 O 31 0 0 3 0 g o -3 30 o -3 31 R. R2/31 0 0 - R3 R3+3R2 0 1 -( o -3 silo 0 0 0 xi > x2 іх3 o - O x3 E IR X2 - сх. is if then eigenector - 23 Va 11

(ii) eigencector corresponding to dz=2+31 Similarly part il į - replace then eigen vector is OT 3 =